projectile motion - interesting question (1 Viewer)

[Porcia]

Could someone please help me solve this question

A ball is projected so that its horizontal range is 45 metres and it passes through a point 22.5 metres horizontally from and 11.25 metres vertically above the point of projection. Find the angle of projection and speed thereof.

 

[rama_v]

Could someone please help me solve this question

A ball is projected so that its horizontal range is 45 metres and it passes through a point 22.5 metres horizontally from and 11.25 metres vertically above the point of projection. Find the angle of projection and speed thereof.

[I'm guessing here]
Sub the point (22.5, 11.25) into the cartesian equation and then sub the point (45, 0) into the same cartesian equation, and simultaneously solve. Oh and you might need the identity
sec²@ = 1 + tan²@

 

[Porcia]

but that makes the cartesian equation too big to handle

 

[rama_v]

but that makes the cartesian equation too big to handle

You can also use the range formula.

 

[Riviet]

Let X be initial horizontal speed and Y be initial vertical speed.
For simplicity, let the ball be projected from ground level, and derive the equations for horizontal and vertical displacement as normal:
x=Xt and y=-gt² + Yt
From the x equation, t=x/X
Substitute into y equation to get:
y=-gx²/X² + Yx/X (1)

it passes through a point 22.5 metres horizontally from and 11.25 metres vertically above the point of projection

- implies that the particle passes through (22.5 , 11.25) so substitute this into (1) to get an equation in X and Y

horizontal range is 45 metres

- implies that when y=0, x=45 so substitute this into (1) for a second equation in X and Y.
Now solve these two equations simultaneously, find X and Y, and use tan@=Y/X to find @. To find speed of projection, use Pythagoras' Theorem. Hope that helps.

P.S Take g=10 if you want numerical answers. Take Shifty's advice too.

 

[_ShiFTy_]

.
y = 0, when x = 22.5 and y = 11.25

You can say this because you are projecting it from ground level, noting that 22.5 is half of 45 meaning that y = 11.25 must be the maximum height

 

[followme]

is it 21.2m/s at 45 deg ? if g=10

 

[SoulSearcher]

Yes, that seems to be the correct answer for the particular question.